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JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Binomial Theorem

Let

C_{r}

denote the coefficient of

x^{r}

in the binomial expansion of

(1 + x)^{n}, n \in N, 0 \leq r \leq n

.

If

P_{n} = C_{0} - C_{1} + \frac{2 ^{2}}{3} C_{2} - \frac{2 ^{3}}{4} C_{3} + \dots .. + \frac{( - 2 ) ^{n}}{n + 1} C_{n}

, then the value of

\sum_{n = 1}^{25} \frac{1}{P _{2 n}}

equals.

(A)580
(B)525
(C)650
(D)675✓ Correct

Answer: (D)

Solution

Using

\frac{C _{r}}{r + 1} = \frac{1}{n + 1} (r + 1 n + 1)

,

P_{n} = r = 0 \sum n \frac{( - 2 ) ^{r}}{r + 1} C_{r} = \frac{- 1}{2 ( n + 1 )} r = 0 \sum n (r + 1 n + 1) (- 2)^{r + 1} = \frac{1 - ( - 1 ) ^{n + 1}}{2 ( n + 1 )} .

For an even index

2 n

this gives

P_{2 n} = \frac{2}{2 ( 2 n + 1 )} = \frac{1}{2 n + 1}

, so

\frac{1}{P _{2 n}} = 2 n + 1

. Therefore

n = 1 \sum 25 \frac{1}{P _{2 n}} = n = 1 \sum 25 (2 n + 1) = 3 + 5 + \dots + 51 = \frac{25}{2} (3 + 51) = 675.

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Single correct option

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